JL=83. if Jk=5x-4 & KL=3x-1. <br> Then X=?

What is a quick and easy way to remember explicit and recursive formulas?

Oliga [24]

I always found derivation to be helpful in remembering. Since this question is tagged as at the middle school level, I assume you've only learned about arithmetic and geometric sequences.

First, remember what these names mean. An arithmetic sequence is a sequence in which consecutive terms are increased by a fixed amount; in other words, it is an additive sequence. If $a_n$ is the $n$ th term in the sequence, then the next term $a_{n+1}$ is a fixed constant (the common difference $d$ ) added to the previous term. As a recursive formula, that's

$a_{n+1}=a_n+d$

This is the part that's probably easier for you to remember. The explicit formula is easily derived from this definition. Since $a_{n+1}=a_n+d$ , this means that $a_n=a_{n-1}+d$ , so you plug this into the recursive formula and end up with

$a_{n+1}=(a_{n-1}+d)+d=a_{n-1}+2d$

You can continue in this pattern, since every term in the sequence follows this rule:

$a_{n+1}=a_{n-1}+2d$
$a_{n+1}=(a_{n-2}+d)+2d$
$a_{n+1}=a_{n-2}+3d$
$a_{n+1}=(a_{n-3}+d)+3d$
$a_{n+1}=a_{n-3}+4d$

and so on. You start to notice a pattern: the subscript of the earlier term in the sequence (on the right side) and the coefficient of the common difference always add up to $n+1$ . You have, for example, $(n-2)+3=n+1$ in the third equation above.

Continuing this pattern, you can write the formula in terms of a known number in the sequence, typically the first one $a_1$ . In order for the pattern mentioned above to hold, you would end up with

$a_{n+1}=a_1+nd$

or, shifting the index by one so that the formula gives the $n$ th term explicitly,

$a_n=a_1+(n-1)d$

Now, geometric sequences behave similarly, but instead of changing additively, the terms of the sequence are scaled or changed multiplicatively. In other words, there is some fixed common ratio $r$ between terms that scales the next term in the sequence relative to the previous one. As a recursive formula,

$a_{n+1}=ra_n$

Well, since $a_n$ is just the term after $a_{n-1}$ scaled by $r$ , you can write

$a_{n+1}=r(ra_{n-1})=r^2a_{n-1}$

Doing this again and again, you'll see a similar pattern emerge:

$a_{n+1}=r^2a_{n-1}$
$a_{n+1}=r^2(ra_{n-2})$
$a_{n+1}=r^3a_{n-2}$
$a_{n+1}=r^3(ra_{n-3})$
$a_{n+1}=r^4a_{n-3}$

and so on. Notice that the subscript and the exponent of the common ratio both add up to $n+1$ . For instance, in the third equation, $3+(n-2)=n+1$ . Extrapolating from this, you can write the explicit rule in terms of the first number in the sequence:

$a_{n+1}=r^na_1$

or, to give the formula for $a_n$ explicitly,

$a_n=r^{n-1}a_1$

6 0

4 years ago

Adina is constructing a line perpendicular to YJ . She has already constructed two arcs as shown. She moves her compass point to

mestny [16]

It has to be wider than the first two arcs or else you will just get two circles side by side with point c in common. I usually measure it to be the length of a point between the c and j. It is a good point to get the arcs to intersect.

8 0

4 years ago

Read 2 more answers

Pricilla currently reads 10 pages in her book each night. She wants to increase the number of pages by 30%.How many pages will P

7nadin3 [17]

She will read 13 pages each night after the increase

6 0

3 years ago

Select the choice that translates the following verbal phrase correctly to algebra:

natta225 [31]

12y+11 is the correct expression.

Product=result of multiplication
Increased= addition
The product of 12 and y increased by 11 is: 12y+11

5 0

3 years ago

How to factor: 15x^2y^3 + 10xy^2z

eduard

Since 5 is a factor of 10 and 15 (by guess and check with factors of 10 and seeing if they're factors of 15), we get 5(3x^2y^3+2xy^2z). Next, y^2 is in both - remember that when multiplying exponents, you add them up, so we end up with 5y^2(x^2y+2xz). Lastly, x (aka x^1) is in both, so we factor that out to end up with 5y^2x(xy+2z)

5 0

3 years ago

JL=83. if Jk=5x-4 & KL=3x-1. Then X=?